Question

我试图运行书中的代码＆＃34; Scala中的函数编程＆＃34;这似乎是为旧版本的scala（从here下载）。 tmp.scala：

sealed trait Option[+A]
case class Some[+A](get: A) extends Option[A] 
case object None extends Option[Nothing]

trait Option[+A] {
    /* returns None if None, or function applied to the some object */
    def map[B](f: A => B): Option[B] = this match {
      case None => None
      case Some(a) => Some(f(a))
    }
}

这引发的错误是：

$ scala
Welcome to Scala 2.12.0-20161021-070700-8684ae8 (OpenJDK 64-Bit Server VM, Java 1.8.0_112).
scala> :load tmp.scala

tmp.scala:17: error: pattern type is incompatible with expected type;
 found   : None.type
 required: Option[A]
             case None => None
                  ^
tmp.scala:17: error: type mismatch;
 found   : None.type
 required: Option[B]
             case None => None
                          ^
tmp.scala:18: error: constructor cannot be instantiated to expected type;
 found   : Some[A(in class Some)]
 required: Option[A(in trait Option)]
             case Some(a) => Some(f(a))
                  ^
tmp.scala:18: error: type mismatch;
 found   : Some[B]
 required: Option[B]
             case Some(a) => Some(f(a))
                                 ^

我尝试了各种各样的代码，但无济于事，似乎由于过时的语法而无法正确检测子类？

Answer 1

使用粘贴内容的:paste file.scala代替:load file.scala来解释每一行。

Scala子类模式匹配

1 个答案: